The HTTP protocol operates at which layer?

Find an overall minimum two-level circuit (corresponding to sum of products expressions) using multiple AND and one multi-input

Goryan [66]

Answer:

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4 0

3 years ago

This part of the electromagnetic spectrum can be used to generate power.

timurjin [86]

A. Gamma Rays is the answer.

6 0

3 years ago

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In the reaction __H2 + O2 → 2H2O, what coefficient should be placed in front of H2 to balance the reaction?

malfutka [58]

The answer is: [B]: " 2 " .
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Explanation:
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Given the chemical equation:
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 ?  H₂ + O₂ → 2 H₂O ;
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 → We are asked, "What coefficient, if any — should be put in front of the: " H₂ " ;
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(which is on the "left-hand side" side of the chemical equation given— the "reactants") ;

→ to get a balanced chemical equation?
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→ Let us examine the "right-hand side" of this chemical equation—the product(s). In this case, the "product" given is: " 2 H₂O " .

So, on the "right hand side", we have:
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1) 4 "H's" → {Two "H₂" 's = 2 * 2 = "4 H's"} ; and:
2) 2 "O's" → { Two "O's").
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So, the left-hand side should have:
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1) 4 "H's" ; and:
2) 2 "O's" ;
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Let us examine the left-hand side (the "reactants").
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"  ?  H₂ + O₂ " ;
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On the left-hand side, we already have:
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1) " 2 O's " ; → one "O₂" = "2 O's" ; and:
2) "2 H's " ; → one "H₂" .

Now, we would need:
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A "total of "4 H's". Is there any number we could put as a coefficient on the other reactant, which happens to be: "H₂" ; to make a total of "4 H's" ?

{Note: There are only these TWO (2) reactants in this chemical equation.}.

 → The "H₂" ; as it stands alone, is insufficient—since that would be only "2 H's".
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→ Thus, we can rule out: "Answer choices: [A] and [D]."
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Note:
→ Choice [A]: "1" ; The coefficient, "1"; is generally not (never?) used; and basically would function as the same as:
→ Choice: [D]: "no coefficient is needed".
→ Choices [A] & [D]: would leave us with only "2 H's" on the "reactants side" (i.e. "left-hand side of the equation"; and we need FOUR ("4 H's").
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Since we are given: "H₂" ; what coefficient could we put in front of this to get: "4 H's" ? (4÷2 =2). So we could put a "2" in front of the "H₂" ; to get:
"4 H's". The coefficient, "2" , corresponds directly with:
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→ Answer choice: [B]: "2" .
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{ NOTE: The remaining answer choice, [C], which is, "3" ; is INCORRECT; since 3 "H₂'s" would be "6 H's" ; (since 3 * "2 H's" equal "6 H's") ; which is too many "H's" → We need exactly "4 H's".}.
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So, the correctly balanced equation is:

2 H₂ + O₂ → 2 H₂O ;
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→ The coefficient that goes before the "H₂" ; to make this chemical equation balanced, is: "2" .
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The correct answer is: [B]: "2" .

→ The coefficient that goes before the "H₂" is: "2" .
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Hope this answer—and {lengthy} explanation—is of help!
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8 0

3 years ago

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Write an algorithm that read a number from the user “assuming the number entered is always between 1-100”, when the number enter

AveGali [126]

import sys

number = int (input ("Enter a number between 1 to 100 : "))

x = 0

y = 0

if number == 0:

sys.exit()

if number > 50:

print("Increment x value")

x=x+1

else:

print("Increment y value")

y=y+1

if x > y:

print("Most of your numbers are greater than 50")

else

print("Most of your numbers are less than or equals to 50")

Explanation:

In this algorithm the User is asked to input the number.

If the input number is 0 then the user will exit the algorithm

If the input number is greater than 50 then X variable is incremented by 1

If the input number is less than 50 then the variable Y is incremented by 1

Then X and Y values are compared,if the value of x is greater than y then

and print command will print Most of your numbers are greater than 50

If the value of Y is greater than X then print command will print Most of your numbers are less than or equal to 50

6 0

3 years ago

You have a K-Way set associative cache with following parameters

MaRussiya [10]

Answer:

Following are the solution to the given question:

Explanation:

The size of cache memory $= 2048 (2^{11})$

The block size $=16 (2^4)$

Therefore, $\frac{2048}{16}=128 (2^7)$ are blocks.

The collection consists of 4 blocks. Therefore, $\frac{128}{4} =32$

are sets. Because the size of its cache block is 16, the block offset comprises 4 bits. There would be 5 bits inset offset since there are 32 caches. An Address contains 32 bits, while the space of the tags contains $32- (4+5)=23$ bits.

6 0

3 years ago